By Am-Gm inequality: \(abc+abc+1\ge3\sqrt[3]{a^2b^2c^2}=\dfrac{3abc}{\sqrt[3]{abc}}\ge\dfrac{9abc}{a+b+c}\)

So we need to prove : \(a^2+b^2+c^2+\dfrac{9abc}{a+b+c}\ge2\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^3+b^3+c^3+3abc\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\)( schur level 3 inequality ) ( proved)

Equality occurs for a=b=c=1 

c2 : Diriclet theorem \(\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\) ( we can assume that)\(\Rightarrow c\left(a-1\right)\left(b-1\right)\ge0\)

\(\Leftrightarrow abc\ge ac+bc-c\) . So we need to prove 

\(a^2+b^2+c^2+2\left(ac+bc-c\right)+1\ge2\left(ab+bc+ca\right)\)

\(\Leftrightarrow\left(a-b\right)^2+\left(c-1\right)^2\ge0\) (true)